C++ auto keyword. Why is it magic?

Question

From all the material I used to learn C++, auto has always been a weird storage duration specifier that didn't serve any purpose. But just recently, I encountered code that used it as a type name in and of itself. Out of curiosity I tried it, and it assumes the type of whatever I happen to assign to it!

Suddenly STL iterators and, well, anything at all that uses templates is 10 fold easier to write. It feels like I'm using a 'fun' language like Python.

Where has this keyword been my whole life? Will you dash my dreams by saying it's exclusive to visual studio or not portable?

Answer 1

It's just taking a generally useless keyword and giving it a new, better functionality. It's standard in C++11, and most C++ compilers with even some C++11 support will support it.

Answer 2

This functionality hasn't been there your whole life. It's been supported in Visual Studio since the 2010 version. It's a new C++11 feature, so it's not exclusive to Visual Studio and is/will be portable. Most compilers support it already.

Answer 3

It's not going anywhere ... it's a new standard C++ feature in the implementation of C++11. That being said, while it's a wonderful tool for simplifying object declarations as well as cleaning up the syntax for certain call-paradigms (i.e., range-based for-loops), don't over-use/abuse it :-)

Answer 4

auto was a keyword that C++ "inherited" from C that had been there nearly forever, but virtually never used because there were only two possible conditions: either it wasn't allowed, or else it was assumed by default.

The use of auto to mean a deduced type was new with C++11.

At the same time, auto x = initializer deduces the type of x from the type f initializer the same way as template type deduction works for function templates. Consider a function template like this:

template<class T>
int whatever(T t) { 
    // point A
};

At point A, a type has been assigned to T based on the value passed for the parameter to whatever. When you do auto x = initializer;, the same type deduction is used to determine the type for x from the type of initializer that's used to initialize it.

This means that most of the type deduction mechanics a compiler needs to implement auto were already present and used for templates on any compiler that even sort of attempted to implement C++98/03. As such, adding support for auto was apparently fairly easy for essentially all the compiler teams--it was added quite quickly, and there seem to have been few bugs related to it either.

When this answer was originally written (in 2011, before the ink was dry on the C++ 11 standard) auto was already quite portable. Nowadays, it's thoroughly portable among all the mainstream compilers. The only obvious reasons to avoid it would be if you need to write code that's compatible with a C compiler, or you have a specific need to target some niche compiler that you know doesn't support it (e.g., a few people still write code for MS-DOS using compilers from Borland, Watcom, etc., that haven't seen significant upgrades in decades). If you're using a reasonably current version of any of the mainstream compilers, there's no reason to avoid it at all though.

Answer 5

For variables, specifies that the type of the variable that is being declared will be automatically deduced from its initializer. For functions, specifies that the return type is a trailing return type or will be deduced from its return statements (since C++14).

Syntax

auto variable initializer   (1) (since C++11)

auto function -> return type    (2) (since C++11)

auto function   (3) (since C++14)

decltype(auto) variable initializer (4) (since C++14)

decltype(auto) function (5) (since C++14)

auto :: (6) (concepts TS)

cv(optional) auto ref(optional) parameter   (7) (since C++14)

Explanation

1) When declaring variables in block scope, in namespace scope, in initialization statements of for loops, etc., the keyword auto may be used as the type specifier. Once the type of the initializer has been determined, the compiler determines the type that will replace the keyword auto using the rules for template argument deduction from a function call (see template argument deduction#Other contexts for details). The keyword auto may be accompanied by modifiers, such as const or &, which will participate in the type deduction. For example, given const auto& i = expr;, the type of i is exactly the type of the argument u in an imaginary template template<class U> void f(const U& u) if the function call f(expr) was compiled. Therefore, auto&& may be deduced either as an lvalue reference or rvalue reference according to the initializer, which is used in range-based for loop. If auto is used to declare multiple variables, the deduced types must match. For example, the declaration auto i = 0, d = 0.0; is ill-formed, while the declaration auto i = 0, *p = &i; is well-formed and the auto is deduced as int.

2) In a function declaration that uses the trailing return type syntax, the keyword auto does not perform automatic type detection. It only serves as a part of the syntax.

3) In a function declaration that does not use the trailing return type syntax, the keyword auto indicates that the return type will be deduced from the operand of its return statement using the rules for template argument deduction.

4) If the declared type of the variable is decltype(auto), the keyword auto is replaced with the expression (or expression list) of its initializer, and the actual type is deduced using the rules for decltype.

5) If the return type of the function is declared decltype(auto), the keyword auto is replaced with the operand of its return statement, and the actual return type is deduced using the rules for decltype.

6) A nested-name-specifier of the form auto:: is a placeholder that is replaced by a class or enumeration type following the rules for constrained type placeholder deduction.

7) A parameter declaration in a lambda expression. (since C++14) A function parameter declaration. (concepts TS)

Notes Until C++11, auto had the semantic of a storage duration specifier. Mixing auto variables and functions in one declaration, as in auto f() -> int, i = 0; is not allowed.

For more info : http://en.cppreference.com/w/cpp/language/auto